Solved by me. Hint: The problem involves two prime divisors with their difference being 2.

http://thuanan.net/toan-d%E1%BA%A7u-tu%E1%BA%A7n/

what is answer?

The answer is $n=18$. First $\left\{ {1,3} \right\} \to 2$；$\left\{ {5,7} \right\} \to 2$；$\left\{ {9,11} \right\} \to 2$；$\left\{ {13,15} \right\} \to 2$. $\left\{ {2,2,2,2,2,4,8} \right\} \to 2$，now the rest numbers are $2,6,10,12,14,16,17,18$.. Next $17 + 2 = 19$；$19 + 10 = 29$；$29 + 14 = 43$；$43 + 16 = 59$；$59 + 12 = 71$；$71 + 2 = 73$；$73 + 6 = 79$；$79 + 18 = 97$,we done.

yunxiu wrote: The answer is $n=18$. First $\left\{ {1,3} \right\} \to 2$；$\left\{ {5,7} \right\} \to 2$；$\left\{ {9,11} \right\} \to 2$；$\left\{ {13,15} \right\} \to 2$. $\left\{ {2,2,2,2,2,4,8} \right\} \to 2$，now the rest numbers are $2,6,10,12,14,16,17,18$.. Next $17 + 2 = 19$；$19 + 10 = 29$；$29 + 14 = 43$；$43 + 16 = 59$；$59 + 12 = 71$；$71 + 2 = 73$；$73 + 6 = 79$；$79 + 18 = 97$,we done. thank you, but full solution?

Denote $S$ is the sum of the numbers on blackboard, then we have ①even+even to $2$, ${S_1} \to {S_2} \leqslant {S_1} - 2$; ②odd+odd to $2$, ${S_1} \to {S_2} \leqslant {S_1} - 2$; ③odd+ even to a odd prime, ${S_1} \to {S_2} \leqslant {S_1}$. And above all ${S_2} \equiv {S_1}(\bmod 2)$, so $\sum\limits_{k = 1}^n k = \frac{{n(n + 1)}}{2} \equiv 97(\bmod 2)$, so $n \equiv 1,2(\bmod 4)$. Because $97 \leqslant \frac{{n(n + 1)}}{2}$, we have $n \geqslant 14$. For $n = 14$, $\frac{{n(n + 1)}}{2} = 105 = 97 + 8$, and we have to change $7$ odd numbers to one odd nember, so we have to do ② $3$ times, so $S \to \leqslant S - \left[ {(1 + 3 + 5 + 7 + 9 + 11) - 2 \cdot 3} \right] \leqslant 105 - 30 < 97$. For $n = 17$, $\frac{{n(n + 1)}}{2} = 153 = 97 + 56$, and we have to change $9$ odd numbers to one odd nember, so we have to do ② $4$ times, so $S \to \leqslant S - \left[ {(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15) - 2 \cdot 4} \right] \leqslant 153 - 56 = 97$. Next we have $5$’s $2$, and $4,6,8,10,12,14,16$ and $17$, and ${S_1} \to {S_2} = {S_1}$ for every times.Between $17,97$ have only $5$ twin primes $17,19;29,31;41,43;59,61;71,73$, we must use $5$’s $2$ for them. But we also need $3$’s $10$,because $29 - 19 = 10$; $41 - 31 = 10$;$71 - 61 = 10$, contradiction. So the answer is $n=18$.

hi, your solution is right and clever but there are some details you didn't mention I just tell them for others I'm sure you're aware of them, $n=15,16$ cant be the answer because in all of 3 moves the number of odd numbers doesn't change mod 2 and odd numbers$1,3,5,7,9,11$ all become 2 after some move it can take more than one that's the reason we can reduce them from $S$.